이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "friend.h"
#include "bits/stdc++.h"
#define pb(x) push_back(x)
using namespace std;
const int N = 1005;
vector<int> adj[N];
void add_edge(int u, int v){
adj[u].pb(v);
adj[v].pb(u);
}
int dp[N][2];
void dfs(int nd, int pr, int val[]){
dp[nd][1] = val[nd];
dp[nd][0] = 0;
for (auto c: adj[nd]){
if (c == pr)
continue;
dfs(c, nd, val);
dp[nd][1] += dp[c][0];
dp[nd][0] += max(dp[c][0], dp[c][1]);
}
}
int vis[N], col[N], matched[N];
void dfs0(int nd, vector<int>&L, vector<int> &R){
vis[nd] = 1;
if (!col[nd]) L.pb(nd);
else R.pb(nd);
for (auto to: adj[nd])
if (!vis[to]){
col[to] = col[nd] ^ 1;
dfs0(to, L, R);
}
}
bool dfs1(int nd){
if (vis[nd])
return 0;
vis[nd] = 1;
for (auto x: adj[nd]){
if (matched[x] == -1 or dfs1(matched[x])){
matched[x] = nd;
return 1;
}
}
return 0;
}
int findSample(int n,int arr[],int host[],int protocol[]){
for (int i = 0; i < n; i++)
adj[i].clear();
bool subtask1 = (n <= 10);
bool subtask2 = true, subtask3 = true;
bool subtask4 = true, subtask5 = true;
// Build graph
for (int i = 1; i < n; i++){
int v = host[i], p = protocol[i];
subtask4 &= (p == 0);
subtask2 &= (p == 1);
subtask3 &= (p == 2);
subtask5 &= (p <= 1);
if (p == 0)//IAmYourFriend
add_edge(v, i);
else if(p == 1){//MyFriendsAreYourFriends
for (auto to: adj[v])
add_edge(to, i);
}
else{//WeAreYourFriends
for (auto to: adj[v])
add_edge(to, i);
add_edge(v, i);
}
}
int answer = 0;
if (subtask1){
for (int mask = 0; mask < (1<<n); mask++){
bool bad = 0;
int sum = 0;
for (int i = 0; i < n; i++)
if (mask>>i&1){
sum += arr[i];
for (auto to: adj[i])
bad |= (mask>>to&1);
}
if (!bad)
answer = max(answer, sum);
}
}
else if(subtask2){
for (int i = 0; i < n; i++)
answer += arr[i];
}
else if (subtask3){
for (int i = 0; i < n; i++)
answer = max(answer, arr[i]);
}
else if (subtask4){
dfs(0, -1, arr);
answer = max(dp[0][0], dp[0][1]);
}
else if(subtask5){
for (int i = 0; i < n; i++)
vis[i] = col[i] = 0;
int MM = 0;
for (int i = 0; i < n; i++)
if (!vis[i]){
vector<int> L, R;
dfs0(i, L, R);
// Kuhn's algorithm
for (auto x: R)
matched[x] = -1;
for (auto x: L){
for (auto y: L)
vis[y] = false;
dfs1(x);
}
for (auto x: R)
MM += (matched[x] != -1);
}
answer = n - MM;
}
return answer;
}
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